Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 427521, 12 pages
http://dx.doi.org/10.1155/2013/427521
Research Article
Numerical Solution of Higher Order Boundary Value Problems
Shahid S. Siddiqi1 and Muzammal Iftikhar1,2
1
2
Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan
Department of Mathematics, University of Education, Okara Campus, Okara 56300, Pakistan
Correspondence should be addressed to Muzammal Iftikhar; miftikhar@hotmail.com
Received 27 December 2012; Revised 5 April 2013; Accepted 6 April 2013
Academic Editor: Hossein Jafari
Copyright © 2013 S. S. Siddiqi and M. Iftikhar. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
The aim of this paper is to use the homotopy analysis method (HAM), an approximating technique for solving linear and nonlinear
higher order boundary value problems. Using HAM, approximate solutions of seventh-, eighth-, and tenth-order boundary value
problems are developed. This approach provides the solution in terms of a convergent series. Approximate results are given for
several examples to illustrate the implementation and accuracy of the method. The results obtained from this method are compared
with the exact solutions and other methods (Akram and Rehman (2013), Farajeyan and Maleki (2012), Geng and Li (2009), Golbabai
and Javidi (2007), He (2007), Inc and Evans (2004), Lamnii et al. (2008), Siddiqi and Akram (2007), Siddiqi et al. (2012), Siddiqi
et al. (2009), Siddiqi and Iftikhar (2013), Siddiqi and Twizell (1996), Siddiqi and Twizell (1998), Torvattanabun and Koonprasert
(2010), and Kasi Viswanadham and Raju (2012)) revealing that the present method is more accurate.
1. Introduction
Higher order boundary value problems occur in the study of
fluid dynamics, astrophysics, hydrodynamic, hydromagnetic
stability, astronomy, beam and long wave theory, induction
motors, engineering, and applied physics. The boundary
value problems of higher order have been examined due to
their mathematical importance and applications in diversified applied sciences.
The seventh-order boundary value problems generally
arise in modeling induction motors with two rotor circuits.
The induction motor behavior is represented by a fifth-order
differential equation model. This model contains two stator
state variables, two rotor state variables, and one shaft speed.
Normally, two more variables must be added to account
for the effects of a second rotor circuit representing deep
bars, a starting cage, or rotor distributed parameters. To
avoid the computational burden of additional state variables
when additional rotor circuits are required, model is often
limited to the fifth-order and rotor impedance is algebraically
altered as function of rotor speed under the assumption
that the frequency of rotor currents depends on rotor speed.
This approach is efficient for the steady state response with
sinusoidal voltage, but it does not hold up during the transient
conditions, when rotor frequency is not a single value. So, the
behavior of such models shows up in the seventh order [1].
Chandrasekhar [2] investigated that when an infinite
horizontal layer of fluid is heated from below and is subject
to rotation, the instability sets in. When this instability sets in
as overstability, it is represented by an eighth-order ordinary
differential equation. If an infinite horizontal layer of fluid
is heated from below, with the assumption that a uniform
magnetic field is applied as well across the fluid in the same
direction as gravity and the fluid is subject to the action of
rotation, the instability sets in. When this instability sets in as
ordinary convection, it is modeled by tenth-order boundary
value problem.
Siddiqi and Iftikhar used the variation of parameter
method for solving the seventh-order boundary value problems in [3]. Liu and Wu [4] give the general differential
quadrature rule (GDQR) for the solution of eighth-order
differential equation. Explicit weighting coefficients are formulated to implement the GDQR for eighth-order differential equations. Siddiqi and Akram [5] used nonic spline and
nonpolynomial spline technique for the numerical solution
of eighth-order linear special case boundary value problems.
These have also been proven to be second order convergent.
2
Siddiqi and Twizell [6] presented the solution of eighth-order
boundary value problem using octic spline. Inc and Evans
[7] presented the solutions of eighth-order boundary value
problems using Adomian decomposition method. Golbabai
and Javidi [8] used homotopy perturbation method (HPM)
to solve eighth-order boundary value problems. Recently,
Akram and Rehman presented the numerical solution of
eighth-order boundary value problems using the reproducing
Kernel space method [9]. Geng and Li [10] construct a reproducing Kernel space and solve a class of linear tenth-order
boundary value problems using reproducing Kernel method.
Siddiqi et al. [11] used the variational iteration technique
for the solution of tenth-order boundary value problem.
Siddiqi and Akram [12] presented the numerical solutions of
the tenth-order linear special case boundary value problems
using eleventh degree spline. Siddiqi and Twizell [13] presented the solutions of tenth-order boundary value problems
using tenth degree spline, where some unexpected results, for
the solution and higher order derivatives, were obtained near
the boundaries of the interval. Lamnii et al. [14] developed
a spline collocation method using spline interpolants and
analyzed the approximating solutions of some general linear
boundary value problems. Domairry and Nadim in [15]
compared the HAM and HPM in solving nonlinear heat
transfer equation. HAM is employed to compute approximate
solution of the system of differential equations governing the
problem [16] and also used to detect the fin excellency of
convective straight fins with temperature-dependent thermal
conductivity in [17]. Moghimi et al. applied HAM to solve
MHD Jeffery-Hamel flows in nonparallel walls [18]. Farajeyan
and Maleki [19] used nonpolynomial spline in off-step points
to solve special tenth order linear boundary value problems.
Khan and Hussain in 2011 applied Laplace decomposition
method (LDM) to nonlinear Blasius flow equation to obtain
series solutions [20]. Khan and Gondal [21] constructed a
new method for the solution of Abel’s type singular integral
equations. The two-step Laplace decomposition algorithm
(TSLDA) makes the calculation much simpler.
Khan et al. [22] proposed a method which efficiently
finds exact solution and is used to solve nonlinear Volterra
integral equations. Khan et al. [23] proposed the coupling
of homotopy perturbation and Laplace transformation for
solving system of partial differential equations. Nadeem et
al. [24] described the stagnation point flow of a viscous fluid
towards a stretching sheet and obtained an analytical solution
of the boundary layer equation by HAM.
Recently, Shaban et al. [25] presented modification of the
HAM for solving nonlinear boundary value problems. Arqub
and El-Ajou [26] investigated the accuracy of the HAM for
solving the fractional order problem of the spread of a disease
in a population. In [27], Russo and Van Gorder discussed the
application of HAM to general nonlinear Klein-Gordon type
equations.
In the present paper, the seventh-, eighth-, and tenthorder boundary value problems are solved using the homotopy analysis method (HAM). The following seventh-,
eighth-, and tenth-order boundary value problems are considered:
𝑢(𝑚) (𝑥) = 𝑓 (𝑥, 𝑢 (𝑥)) , 𝑎 ≤ 𝑥 ≤ 𝑏,
Abstract and Applied Analysis
𝑢(𝑖) (𝑎) = 𝐴 𝑖 ,
𝑢(𝑗) (𝑏) = 𝐵𝑗 ,
(1)
where for 𝑚 = 7, 𝑖 = 0, 1, 2, . . . , 𝑚 − 4 and 𝑗 = 0, 1, . . . , 𝑚 − 5;
for 𝑚 = 8, 𝑖 = 𝑗 = 0, 1, 2, . . . , 𝑚 − 5, and for 𝑚 = 10, 𝑖 = 𝑗 =
0, 1, 2, . . . , 𝑚 − 6. 𝐴 𝑖 ’s and 𝐵𝑗 ’s are finite real constants. Also,
𝑓(𝑥, 𝑢(𝑥)) is a continuous function on [𝑎, 𝑏].
2. Homotopy Analysis Method
Liao was the first to apply homotopy analysis method (HAM)
[28–31]. This is a general analytic approach to get series solutions of nonlinear equations, including algebraic equations,
ordinary differential equations, partial differential equations,
differential-integral equations, differential-difference equation, and coupled equations of them. For a given nonlinear
differential equation
𝑁 [𝑢 (𝑥)] = 0,
𝑥 ∈ Θ,
(2)
where 𝑁 is a nonlinear operator and 𝑢(𝑥) is an unknown
function, Liao constructed a one parameter family of equations in the embedding parameter 𝑞 ∈ [0, 1], called the zeroth-order deformation equation
(1 − 𝑞) 𝐿 [𝑈 (𝑥, 𝑞) − 𝑢0 (𝑥)] − 𝑞ℎ𝐻 (𝑥) 𝑁 [𝑈 (𝑥, 𝑞)] = 0,
𝑥 ∈ Θ, 𝑞 ∈ [0, 1] ,
(3)
where ℎ is a nonzero auxiliary parameter, 𝐻(𝑥) is an auxiliary
function, 𝐿 is an auxiliary linear operator, 𝑢0 (𝑥) is an initial
guess, and 𝑈(𝑥, 𝑞) is an unknown function. The homotopy
provides us larger freedom to choose both the auxiliary
linear operator 𝐿 and the initial guess than the traditional
nonperturbation methods, as pointed out by Liao [29, 31]. At
𝑞 = 0 and 𝑞 = 1, we have 𝑈(𝑥, 0) = 𝑢0 (𝑥) and 𝑈(𝑥, 1) = 𝑢(𝑥),
respectively. Thus, as 𝑞 increases from 0 to 1, the solution
𝑈(𝑥, 𝑞) varies from the initial guess 𝑢0 (𝑥) to the solution
𝑢(𝑥). Expanding 𝑈(𝑥, 𝑞) by Taylor series with respect to 𝑞,
(2) becomes
∞
𝑈 (𝑥, 𝑞) = 𝑢0 (𝑥) + ∑ 𝑢𝑚 (𝑥) 𝑞𝑚 ,
(4)
𝑚=1
where
𝑢𝑚 (𝑥) =
󵄨
𝑚
1 𝜕 𝑈 (𝑥, 𝑞) 󵄨󵄨󵄨
󵄨󵄨 .
𝑚!
𝜕𝑞𝑚 󵄨󵄨󵄨𝑞=0
(5)
If the auxiliary linear operator, the initial guess, the auxiliary
parameter ℎ, and the auxiliary function are properly chosen,
series (4) converges at 𝑞 = 1, and then the homotopy series
solution
∞
𝑢 (𝑥) = 𝑢0 + ∑ 𝑢𝑚 (𝑥)
𝑚=1
(6)
Abstract and Applied Analysis
3
must be one of solutions of original equations 𝑁[𝑢(𝑥)] =
0 [29]. Here, 𝑢𝑚 (𝑥) is governed by a linear differential
equation related to the auxiliary linear operator 𝐿. According
to definition (6), the governing equation can be deduced from
the zeroth-order deformation (4). Define the vector
𝑢⃗𝑁 (𝑥) = {𝑢0 (𝑥) , 𝑢1 (𝑥) , . . . , 𝑢𝑁 (𝑥)} .
𝑥 ∈ Θ, 𝑞 ∈ [0, 1] ,
(8)
where
𝑅𝑚 (𝑢⃗𝑚−1 ) =
𝜕𝑚−1 𝑁 (𝑈 (𝑥, 𝑞)) 󵄨󵄨󵄨󵄨
1
󵄨󵄨 ,
󵄨󵄨
𝜕𝑞𝑚−1
(𝑚 − 1)!
󵄨𝑞=0
0,
𝜒𝑚 = {
1,
𝑚 ≤ 1,
𝑚 ≥ 1.
𝑢0 (𝑥) = 𝑥 −
𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻 (𝑥) 𝑅𝑚 (𝑢⃗𝑚−1 ) ,
(9)
𝜕7 𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑞7
(10)
To implement the HAM, several numerical examples are
considered in the following section.
3. Numerical Examples
Example 1. Consider the following seventh-order boundary
value problem:
𝑅𝑚 (𝑢⃗𝑚−1 ) =
𝑢(2) (0) = 0,
𝑢(2) (1) = −4𝑒,
𝜕7 𝑢𝑚−1 (𝑥, 𝑞)
+ 𝑢𝑚−1 (𝑥, 𝑞) ,
𝜕𝑞7
𝑚 ≥ 2.
𝑢𝑚 (𝑥) = 𝜒𝑚 𝑢𝑚−1 (𝑥) + ℎ𝐿−1 [𝑅𝑚 (𝑢⃗𝑚−1 )] .
𝑢0 (𝑥) = 𝑥 −
𝑥3 1
+ (−17 + 6𝑒) 𝑥4
2
2
1
1
+ (27 − 10𝑒) 𝑥5 + (−11 + 4𝑒) 𝑥6 ,
2
2
(11)
𝑢1 (𝑥) =
ℎ
518918400
× (−32691859200 + 32691859200𝑒𝑥
The exact solution of Example 1 is 𝑢(𝑥) = 𝑥(1 − 𝑥)𝑒𝑥 [3].
Using the HAM (3), the zeroth-order deformation is
given by
− 24389164800𝑥 − 8302694400𝑒𝑥 𝑥
− 9081072000𝑥2 + 1037836800𝑒𝑥 𝑥2
− 2335132800𝑥3 + 795242426295𝑥4
(1 − 𝑞) 𝐿 [𝑈 (𝑥, 𝑞) − 𝑢0 (𝑥)]
𝜕7 𝑈 (𝑥, 𝑞)
= 𝑞ℎ𝐻 (𝑥) (
+ 𝑈 (𝑥, 𝑞) + 𝑒𝑥 (35 + 12𝑥 + 2𝑥2 )) .
𝜕𝑞7
(12)
(17)
Consequently, the first few terms of the homotopy series
solution are as follows:
0 ≤ 𝑥 ≤ 1,
𝑢(3) (0) = −3.
(16)
Now, the solution of the 𝑚th-order deformation equations
(15) and (16) for 𝑚 ≥ 1 becomes
𝑢 (1) = 0,
𝑢(1) (1) = −𝑒,
2
+ 𝑢𝑚−1 (𝑥, 𝑞) + 𝑒 (35 + 12𝑥 + 2𝑥 ) ,
𝑚=1
𝑢(1) (0) = 1,
(15)
𝑥
𝑁
𝑢 (0) = 0,
(14)
where
Finally, an 𝑁-th order approximate solution is given by
𝑢(7) (𝑥) = −𝑢 (𝑥) − 𝑒𝑥 (35 + 12𝑥 + 2𝑥2 ) ,
(13)
The linear operator 𝐿 normally consists of the homogeneous
part of nonlinear operator 𝑁, whereas parameter ℎ and
function 𝐻(𝑥) are introduced in order to optimize the initial
guess. Try to choose ℎ in such a way that they get a convergent
series. Under the rule of solution expression (4), the auxiliary
function 𝐻(𝑥) can be chosen as 𝐻(𝑥) = 1. In this way, good
approximations of such problems can be obtained without
having to go up to high order of approximation and without
requiring a small parameter.
Hence, the 𝑚th-order deformation can be given by
𝑅1 (𝑢⃗0 ) =
𝑢 (𝑥) = 𝑢0 + ∑ 𝑢𝑚 (𝑥) .
𝑥3 1
+ (−17 + 6𝑒) 𝑥4
2
2
1
1
+ (27 − 10𝑒) 𝑥5 + (−11 + 4𝑒) 𝑥6 .
2
2
(7)
Differentiating equation (3) 𝑚 times with respect to the
embedded homotopy parameter 𝑞, then setting 𝑞 = 0, and
then finally dividing them by 𝑚!, the 𝑚th-order deformation
equation is obtained as
𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] − ℎ𝐻 (𝑥) 𝑅𝑚 (𝑢⃗𝑚−1 ) = 0,
Now, the initial approximation, 𝑢0 (𝑥), is the solution of
(𝜕7 /𝜕𝑥7 )𝑢 = 0 subject to boundary conditions in (11); that
is,
− 292929440550𝑒𝑥4 − 1230096003906𝑥5
+ 452496853920𝑒𝑥5 + 502831877247𝑥6
− 184994415376𝑒𝑥6 + 518918400𝑥7
(18)
4
Abstract and Applied Analysis
×10−13
0.4
12
0.3
10
8
0.2
6
0.1
4
2
0.2
0.6
0.4
0.8
1
0.2
Figure 1: Comparison of the approximate solution with the exact
solution for problem (11). Dotted line: approximate solution; solid
line: the exact solution.
12
+ 936𝑒𝑥 + 1755𝑥 − 650𝑒𝑥
13
0.8
1
Figure 2: Absolute errors for problem (11).
𝑛
∑ [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)]
12
(24)
𝑚=1
13
−330𝑥 + 120𝑒𝑥 ) .
= 𝑢1 + (𝑢2 − 𝑢1 ) + (𝑢3 − 𝑢2 ) + ⋅ ⋅ ⋅ + (𝑢𝑛 − 𝑢𝑛−1 ) .
(19)
Using (22),
The 𝑁th-order approximation of 𝑢(𝑥) can be expressed by
∞
∑ [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = lim 𝑢𝑛 (𝑥) = 0,
𝑁
𝑢𝑁 (𝑥) = 𝑢0 (𝑥) + ∑ 𝑢𝑚 (𝑥) .
0.6
We have
+ 12870𝑥8 − 429𝑥10 − 2652𝑥11
11
0.4
(20)
𝑚=1
Equation (20) is a family of the approximate solutions to
problem (11) in terms of the convergence control parameter
ℎ.
We choose the value of auxiliary parameter as ℎ = −1
to ensure that the solution series converges. The errors in
absolute values obtained using the present method are compared with those obtained using the variation of parameter
method [3] for Example 1 given in Table 1, which shows that
the present method is quite accurate. Figures 1 and 2 show the
comparison of exact with approximate solution and absolute
errors for Example 1 solution respectively.
Convergence Theorem. In this subsection, one proves that if
the solution series (6) given by HAM is convergent, it must be
an exact solution of the considered problem.
If the series 𝑢0 (𝑥) + ∑∞
𝑚=1 𝑢𝑚 (𝑥) converges, where 𝑢𝑚 (𝑥) is
governed by (17) under the definitions (15) and (16), it must be
an exact solution of problem (11).
𝑛→∞
𝑚=1
(25)
and applying the operator 𝐿, we can write
∞
𝐿 ∑ [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)]
𝑚=1
(26)
∞
= ∑ 𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = 0;
𝑚=1
using the definition (14),
∞
∞
𝑚=1
𝑚=1
∑ 𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻 (𝑥) ∑ 𝑅𝑚 (𝑢⃗𝑚−1 ) , (27)
and since ℎ ≠ 0, 𝐻(𝑥) ≠ 0, we have
∞
∑ 𝑅𝑚 (𝑢⃗𝑚−1 ) = 0.
(28)
𝑚=1
From (15) and (16), the following holds:
∞
∑ 𝑅𝑚 (𝑢⃗𝑚−1 )
Proof. Let the series
𝑚=1
∞
∑ 𝑢𝑚 (𝑥)
(21)
𝑚=0
∞
= ∑(
𝑚=1
𝜕7 𝑢𝑚−1 (𝑥, 𝑞)
+ 𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑞7
be convergent. Then,
+ (1 − 𝜒𝑚 ) 𝑒𝑥 (35 + 12𝑥 + 2𝑥2 ) )
∞
𝑢 (𝑥) = ∑ 𝑢𝑚 (𝑥) ,
(22)
lim 𝑢𝑛 (𝑥) = 0.
(23)
𝑚=0
𝑛→∞
= 0.
This completes the proof.
(29)
Abstract and Applied Analysis
5
Table 1: Comparison of numerical results for Example 1.
𝑥
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Exact solution
0.0000
0.9946
0.1954
0.2835
0.3580
0.4122
0.4373
0.4229
0.3561
0.2214
0.0000
Approximate series solution
0.0000
0.9946
0.1954
0.2835
0.3580
0.4122
0.4373
0.4229
0.3561
0.2214
−1.65159𝐸 − 12
Absolute error present method
0.0000
5.39291𝐸 − 14
4.85167𝐸 − 14
3.92464𝐸 − 14
2.21489𝐸 − 14
3.84137𝐸 − 14
2.10831𝐸 − 13
1.99785𝐸 − 13
3.29736𝐸 − 13
1.77622𝐸 − 12
1.65159𝐸 − 12
Example 2. The following seventh-order nonlinear boundary
value problem is considered:
𝑢(7) (𝑥) = 𝑢 (𝑥) 𝑢󸀠 (𝑥) + 𝑒−2𝑥 (2 + 𝑒𝑥 (𝑥 − 8) − 3𝑥 + 𝑥2 ) ,
0 ≤ 𝑥 ≤ 1,
𝑢 (0) = 1,
𝑢(1) (0) = −2,
𝑢
(2)
(0) = 3,
𝑢 (1) = 0,
𝑢
series. Under the rule of solution expression (4), the auxiliary
function 𝐻(𝑥) can be chosen as 𝐻(𝑥) = 1. In this way, good
approximations of such problems can be obtained without
having to go up to high order of approximation and without
requiring a small parameter.
Hence, the 𝑚th-order deformation can be given by
𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻 (𝑥) 𝑅𝑚 (𝑢⃗𝑚−1 ) ,
1
𝑢(1) (1) = − ,
𝑒
(2)
2
(1) = ,
𝑒
𝑅1 (𝑢⃗0 ) =
The exact solution of Example 2 is 𝑢(𝑥) = (1 − 𝑥)𝑒−𝑥 .
Using the HAM (3), the zeroth-order deformation is
given by
(1 − 𝑞) 𝐿 [𝑈 (𝑥, 𝑞) − 𝑢0 (𝑥)]
𝜕7 𝑈 (𝑥, 𝑞)
𝜕𝑈 (𝑥, 𝑞)
= 𝑞ℎ𝐻 (𝑥) (
+ 𝑈 (𝑥, 𝑞)
7
𝜕𝑞
𝜕𝑞
(31)
6
(−66 + 24𝑒) 𝑥
(30 − 11𝑒) 𝑥
+
.
6𝑒
6𝑒
(34)
𝜕7 𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑞7
+ 𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑢𝑚−1 (𝑥, 𝑞)
,
𝜕𝑞
𝑚 ≥ 2.
𝑢𝑚 (𝑥) = 𝜒𝑚 𝑢𝑚−1 (𝑥) + ℎ𝐿−1 [𝑅𝑚 (𝑢⃗𝑚−1 )] .
(35)
Consequently, the first few terms of the homotopy series
solution are as follows:
Now, the initial approximation, 𝑢0 (𝑥), is the solution of
(𝜕7 /𝜕𝑥7 )𝑢 = 0 subject to boundary conditions in (30); that
is,
3𝑥2 2𝑥3 (36 − 12𝑒) 𝑥4
−
+
2
3
6𝑒
𝑅𝑚 (𝑢⃗𝑚−1 ) =
Now, the solution of the 𝑚th-order deformation equations
(34) for 𝑚 ≥ 1 becomes
+𝑒−2𝑥 (2 + 𝑒𝑥 (−8 + 𝑥) − 3𝑥 + 𝑥2 ) ) .
+
(𝑥, 𝑞)
𝜕7 𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑢
+ 𝑢𝑚−1 (𝑥, 𝑞) 𝑚−1
𝜕𝑞7
𝜕𝑞
+ 𝑒−2𝑥 (2 + 𝑒𝑥 (−8 + 𝑥) − 3𝑥 + 𝑥2 ) ,
(30)
5
(33)
where
𝑢(3) (0) = −4.
𝑢0 (𝑥) = 1 − 2𝑥 +
Absolute error Siddiqi and Iftikhar [3]
0.0000
8.55607𝐸 − 13
9.94041𝐸 − 12
3.52244𝐸 − 11
7.3224𝐸 − 10
1.08769𝐸 − 10
1.29035𝐸 − 10
1.51466𝐸 − 10
2.717974𝐸 − 10
7.48179𝐸 − 10
2.1729𝐸 − 09
(32)
The linear operator 𝐿 normally consists of the homogeneous
part of nonlinear operator 𝑁, whereas parameter ℎ and
function 𝐻(𝑥) are introduced in order to optimize the initial
guess. Try to choose ℎ in such a way that they get a convergent
𝑢0 (𝑥) = 1 − 2𝑥 +
3𝑥2 2𝑥3 (36 − 12𝑒) 𝑥4
−
+
2
3
6𝑒
(−66 + 24𝑒) 𝑥5 (30 − 11𝑒) 𝑥6
+
+
,
6𝑒
6𝑒
𝑢1 (𝑥) =
(36)
ℎ𝑒−2−2𝑥
52929676800
× ( 2274322050𝑒2 − 52929676800𝑒2+𝑥
+ 50655354750𝑒2+2𝑥 + 1654052400𝑒2 𝑥
+ 52929676800𝑒2+𝑥 𝑥 − 102964761900𝑒2+2𝑥 𝑥
6
Abstract and Applied Analysis
+ 413513100𝑒2 𝑥2 + 77740462800𝑒2+2𝑥 𝑥2
1
− 34735100400𝑒2+2𝑥 𝑥3 − 100276694820𝑒2𝑥 𝑥4
0.8
+ 317578525200𝑒1+2𝑥 𝑥4 − 92426387280𝑒2+2𝑥 𝑥4
0.6
+ 168299259192𝑒2𝑥 𝑥5 − 582227931918𝑒1+2𝑥 𝑥5
0.4
2+2𝑥 5
2𝑥 6
1+2𝑥 6
2+2𝑥 6
+ 188997561564𝑒
+ 264649706520𝑒
𝑥 − 72364427100𝑒 𝑥
𝑥 − 87281523925𝑒
0.2
𝑥
0.2
+ 21003840𝑒2+2𝑥 𝑥7 − 9189180𝑒2+2𝑥 𝑥8
0.6
0.8
1
Figure 3: Comparison of the approximate solution with the exact
solution for problem (12). Dotted line: approximate solution; solid
line: the exact solution.
+ 3208920𝑒2+2𝑥 𝑥9 − 2100384𝑒1+2𝑥 𝑥10
+ 160446𝑒2+2𝑥 𝑥10 + 3659760𝑒1+2𝑥 𝑥11
− 1113840𝑒2+2𝑥 𝑥11 − 2864160𝑒1+2𝑥 𝑥12
×10−12
2
+ 1003340𝑒2+2𝑥 𝑥12 + 1306620𝑒1+2𝑥 𝑥13
− 471240𝑒2+2𝑥 𝑥13 − 440640𝑒2𝑥 𝑥14
1.5
+ 69360𝑒1+2𝑥 𝑥14 + 83640𝑒2+2𝑥 𝑥14
2𝑥 15
0.4
1
1+2𝑥 15
+ 969408𝑒 𝑥 − 626688𝑒
𝑥
0.5
+ 99552𝑒2+2𝑥 𝑥15 − 830790𝑒2𝑥 𝑥16
+ 596700𝑒1+2𝑥 𝑥16 − 107100𝑒2+2𝑥 𝑥16
0.2
+ 326700𝑒2𝑥 𝑥17 − 238590𝑒1+2𝑥 𝑥17
0.4
0.6
0.8
1
Figure 4: Absolute errors for problem (12).
+ 43560𝑒2+2𝑥 𝑥17 − 49500𝑒2𝑥 𝑥18
Table 2: Comparison of numerical results for Example 2.
+36300𝑒1+2𝑥 𝑥18 − 6655𝑒2+2𝑥 𝑥18 ) .
(37)
The 𝑁th-order approximation can be expressed by
𝑁
𝑢 (𝑥) = 𝑢0 (𝑥) + ∑ 𝑢𝑀 (𝑥) .
(38)
𝑚=1
Equation (38) is a family of the approximate solutions to
problem (12) in terms of the convergence control parameter
ℎ.
We choose the value of auxiliary parameter as ℎ =
−1 to ensure that the solution series converges. In Table 2,
the exact solution and the series solution of Example 2 are
compared, which shows that the method is quite accurate.
Figure 3 shows the comparison of exact solution with approximate solution, and Figure 4 shows the absolute errors for
Example 2.
Example 3. The following seventh-order linear boundary
value problem is considered:
𝑢(7) (𝑥) = 𝑥𝑢 (𝑥) + 𝑒𝑥 (𝑥2 − 2𝑥 − 6) , 0 ≤ 𝑥 ≤ 1,
𝑢 (0) = 1,
𝑢(1) (0) = 0,
𝑥
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Exact solution
Approximate series
solution
Absolute error
1.0000
0.814354
0.654985
0.518573
0.402192
0.303265
0.219525
0.148976
0.0898658
0.040657
0.0000
1.0000
0.814354
0.654985
0.518573
0.402192
0.303265
0.219525
0.148976
0.0898658
0.040657
−1.29172𝐸 − 15
0.0000
4.15223𝐸 − 14
4.18332𝐸 − 13
1.21736𝐸 − 12
1.95471𝐸 − 12
2.03731𝐸 − 12
1.37063𝐸 − 12
4.66988𝐸 − 13
4.8378𝐸 − 14
6.00561𝐸 − 14
1.29172𝐸 − 15
𝑢(2) (0) = −1,
𝑢(2) (1) = −2𝑒,
𝑢(3) (0) = −2.
(39)
𝑢 (1) = 0,
𝑢(1) (1) = −𝑒,
The exact solution of Example 3 is 𝑢(𝑥) = (1 − 𝑥)𝑒𝑥 [32].
Abstract and Applied Analysis
7
1
present method is quite accurate. In Figure 5, the comparison
of exact solution with approximate solution is shown, and
Figure 6 shows absolute errors for Example 3.
0.8
0.6
Example 4. Consider the following eighth-order boundary
value problem:
0.4
𝑢(8) (𝑥) = −𝑥𝑢 (𝑥) − 𝑒𝑥 (48 + 15𝑥 + 2𝑥3 ) ,
0.2
𝑢 (0) = 0,
0.2
0.4
0.6
1
0.8
Figure 5: Comparison of the approximate solution with the exact
solution for problem (13). Dotted line: approximate solution; solid
line: the exact solution.
0 ≤ 𝑥 ≤ 1,
𝑢 (1) = 0,
𝑢(1) (0) = 1,
𝑢(1) (1) = −𝑒,
𝑢(2) (0) = 0,
𝑢(2) (1) = −4𝑒,
𝑢(3) (0) = −3,
𝑢(3) (1) = −9𝑒.
(40)
×10−12
The exact solution of Example 4 is 𝑢(𝑥) = 𝑥(1 − 𝑥)𝑒𝑥 [5, 7, 9].
Using the HAM (3), the zeroth-order deformation is
given by
8
6
(1 − 𝑞) 𝐿 [𝑈 (𝑥, 𝑞) − 𝑢0 (𝑥)]
4
= 𝑞ℎ𝐻 (𝑥) (
𝜕8 𝑈 (𝑥, 𝑞)
+ 𝑥𝑈 (𝑥, 𝑞)
𝜕𝑞8
(41)
2
+𝑒𝑥 (48 + 15𝑥 + 2𝑥3 ) ) .
0.2
0.4
0.6
0.8
1
Figure 6: Absolute errors for problem (13).
Table 3: Comparison of numerical results for Example 3.
𝑥
Exact
solution
Approximate
series
solution
Absolute
error
present
method
Absolute
error
Siddiqi et al.
[32]
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.0000
0.0994
0.9771
0.9449
0.8950
0.8243
0.7288
0.6041
0.4451
0.2459
0.0000
1.0000
0.0994
0.9771
0.9449
0.8950
0.8243
0.7288
0.6041
0.4451
0.2459
1.20811𝐸 − 11
0.0000
3.41727𝐸 − 13
6.25056𝐸 − 14
1.42442𝐸 − 13
8.83738𝐸 − 14
6.43929𝐸 − 14
1.51812𝐸 − 12
1.47904𝐸 − 12
4.94338𝐸 − 12
5.3817𝐸 − 12
1.20811𝐸 − 11
0.0000
4.6585𝐸 − 13
5.7126𝐸 − 12
2.1299𝐸 − 11
4.6995𝐸 − 11
7.4307𝐸 − 11
8.9219𝐸 − 11
7.9767𝐸 − 11
4.6686𝐸 − 11
1.0960𝐸 − 11
6.9252𝐸 − 16
Following the procedure of the previous example, this
problem is solved using the convergence control parameter
ℎ = −1.
The comparison of the absolute errors obtained by the
present method and the absolute errors obtained by the
method in [32] is given in Table 3, which shows that the
Now, the initial approximation, 𝑢0 (𝑥), is the solution of
(𝜕8 /𝜕𝑥8 )𝑢 = 0 subject to boundary conditions in (40); that
is,
𝑢0 (𝑥) = 𝑥 −
𝑥3 1
1
+ (−36 + 13𝑒) 𝑥4 + (84 − 31𝑒) 𝑥5
2
2
2
1
1
+ (−68 + 25𝑒) 𝑥6 + (19 − 7𝑒) 𝑥7 .
2
2
(42)
The linear operator 𝐿 normally consists of the homogeneous
part of nonlinear operator 𝑁, whereas parameter ℎ and
function 𝐻(𝑥) are introduced in order to optimize the initial
guess. Try to choose ℎ in such a way that they get a convergent
series. Under the rule of solution expression (4), the auxiliary
function 𝐻(𝑥) can be chosen as 𝐻(𝑥) = 1. In this way, good
approximations of such problems can be obtained without
having to go up to high order of approximation and without
requiring a small parameter.
Hence, the 𝑚th-order deformation can be given by
𝐿 [𝑢𝑚 (𝑥) − 𝜒𝑚 𝑢𝑚−1 (𝑥)] = ℎ𝐻 (𝑥) 𝑅𝑚 (𝑢⃗𝑚−1 ) ,
(43)
where
𝑅1 (𝑢⃗0 ) =
𝜕8 𝑢𝑚−1 (𝑥, 𝑞)
+ 𝑥𝑢𝑚−1 (𝑥, 𝑞)
𝜕𝑞8
+ 𝑒𝑥 (48 + 15𝑥 + 2𝑥3 ) ,
𝑅𝑚 (𝑢⃗𝑚−1 ) =
𝜕8 𝑢𝑚−1 (𝑥, 𝑞)
+ 𝑢𝑚−1 (𝑥, 𝑞) ,
𝜕𝑞8
(44)
𝑚 ≥ 2.
8
Abstract and Applied Analysis
Table 4: Comparison of absolute errors for Example 4.
𝑥
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Exact solution
Approximate
series solution
Absolute error
present method
Akram and Rehman
[9]
0.0994654
0.195424
0.28347
0.358038
0.41218
0.437309
0.422888
0.356087
0.0994654
0.195424
0.28347
0.358038
0.41218
0.437309
0.422888
0.356087
3.89966𝐸 − 15
9.45355𝐸 − 14
7.04437𝐸 − 14
4.36873𝐸 − 13
1.28897𝐸 − 13
4.01956𝐸 − 13
2.06929𝐸 − 12
2.65915𝐸 − 12
1.63𝐸 − 10
1.63𝐸 − 09
4.90𝐸 − 09
8.46𝐸 − 09
1.01𝐸 − 08
8.68𝐸 − 09
5.15𝐸 − 09
1.76𝐸 − 09
Siddiqi and Akram
Inc and Evans [7]
[5]
5.62𝐸 − 10
4.88𝐸 − 09
1.37𝐸 − 08
2.29𝐸 − 08
2.71𝐸 − 08
2.38𝐸 − 08
1.49𝐸 − 08
5.54𝐸 − 09
3.73𝐸 − 09
6.61𝐸 − 09
2.33𝐸 − 08
5.17𝐸 − 08
9.76𝐸 − 08
1.78𝐸 − 06
4.12𝐸 − 06
1.83𝐸 − 04
×10−12
0.4
8
0.3
6
0.2
4
0.1
2
0.2
0.4
0.6
0.8
1
0.2
Figure 7: Comparison of the approximate solution with the exact
solution for problem (14). Dotted line: approximate solution; solid
line: the exact solution.
0.4
𝑁
𝑢 (𝑥) = 𝑢0 (𝑥) + ∑ 𝑢𝑚 (𝑥) .
(46)
1
0.3
0.2
0.1
(45)
Consequently, the approximations 𝑢0 , 𝑢1 ,. . . of the homotopy
series solution are obtained.
The 𝑁-th order approximation can be expressed by
0.8
Figure 8: Absolute errors.
Now, the solution of the 𝑚th-order deformation equations
(44) for 𝑚 ≥ 1 becomes
𝑢𝑚 (𝑥) = 𝜒𝑚 𝑢𝑚−1 (𝑥) + ℎ𝐿−1 [𝑅𝑚 (𝑢⃗𝑚−1 )] .
0.6
−1
−0.5
−0.1
0.5
1
−0.2
−0.3
𝑚=1
Equation (46) is a family of the approximate solutions to
problem (14) in terms of the convergence control parameter
ℎ.
We choose the value of auxiliary parameter as ℎ = −1 to
ensure that the solution series converges. For problem (14),
comparison of the results of the present method with the
results of Akram and Rehman [9], Inc and Evans [7], and
Siddiqi and Akram [5] is shown in Table 4. It is observed
that the errors in absolute values of the present method are
better. The comparison of exact solution with approximate
solution is shown in Figure 7, and absolute errors are shown
in Figure 8, respectively.
Example 5. Consider the following boundary value problem:
𝑢(8) (𝑥) = 𝑢 (𝑥) − 8 (2𝑥 cos (𝑥) + 7 sin (𝑥)) ,
𝑢 (−1) = 𝑢 (1) = 0,
−1 ≤ 𝑥 ≤ 1,
Figure 9: Comparison of the approximate solution with the exact
solution for problem (15). Dotted line: approximate solution, solid
line: the exact solution.
𝑢(1) (−1) = 𝑢(1) (1) = 2 sin (1) ,
𝑢(2) (−1) = −𝑢(2) (1) = −4 cos (1) − 2 sin (1) ,
𝑢(3) (−1) = 𝑢(3) (1) = 6 cos (1) − 6 sin (1) .
(47)
The exact solution of problem (15) is 𝑢(𝑥) = (𝑥2 − 1) sin(𝑥)
[5, 6, 9, 14].
Following the procedure of the previous example, this
problem is solved using the convergence control parameter
as ℎ = −1.
Abstract and Applied Analysis
9
×10−13
×10−13
5
3.5
3
4
2.5
3
2
1.5
2
1
1
0.5
−1
0.5
−0.5
1
0.2
0.4
0.6
0.8
1
Figure 12: Absolute errors for problem (16).
Figure 10: Absolute errors for problem (15).
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
−1
0.2
0.4
0.6
0.8
Table 5: Comparison of maximum absolute errors for Problem (15).
Siddiqi
Siddiqi and
Akram and Lamnii et al.
and
Twizell [6]
Rehman [9]
[14]
𝑥 ∈ [𝑥4 , 𝑥𝑘−4 ] Akram [5]
1.89×10−13 4.90 × 10−9
5.01 × 10−9
1.20 × 10−5
1.02 × 10−8
It is observed that the maximum absolute error values
are better than those of Akram and Rehman [9], Lamnii
et al. [14], Siddiqi and Akram [5], and Siddiqi and Twizell
[6] as shown in Table 5. The comparison of exact solution
with approximate solution is shown in Figure 9, and absolute
errors are shown in Figure 10, respectively.
Example 6. Consider the following boundary value problem:
𝑢
(8)
𝑥
(𝑥) = −8𝑒 + 𝑢 (𝑥) ,
𝑢 (0) = 1,
𝑢(2) (0) = −1,
𝑢(1) (1) = −𝑒,
𝑢(3) (0) = −2,
𝑢(2) (1) = −2𝑒.
Figure 13: Comparison of the approximate solution with the exact
solution for problem (17). Dotted line: approximate solution; solid
line: the exact solution.
The exact solution of problem (16) is 𝑢(𝑥) = (1 − 𝑥)𝑒𝑥 [8, 9,
33, 34].
Following the procedure of the previous example, this
problem is solved using the convergence control parameter
as ℎ = −1.
It is observed that the errors in absolute values are better
than those of Akram and Rehman [9], Golbabai and Javidi
[8], He [33], and Torvattanabun and Koonprasert [34] as
shown in Table 6. Figures 11 and 12 show the comparison of
exact solution with approximate solution and absolute errors,
respectively.
Example 7. Consider the following tenth-order boundary
value problem:
= −𝑒𝑥 (89 + 21𝑥 + 𝑥2 − 𝑥3 ) + 𝑥𝑢 (𝑥) ,
𝑢(4) (0) = −3,
𝑢(5) (0) = −4,
1
𝑢(10) (𝑥)
0 < 𝑥 < 1,
𝑢(1) (0) = 0,
0.5
1
Figure 11: Comparison of the approximate solution with the exact
solution for problem (16). Dotted line: approximate solution; solid
line: the exact solution.
Present
method
−0.5
(48)
𝑢 (−1) = 0,
𝑢 (1) = 0,
2
𝑢(1) (−1) = ,
𝑒
𝑢(1) (1) = −2𝑒,
2
𝑢(2) (−1) = ,
𝑒
𝑢(2) (1) = −6𝑒,
−1 < 𝑥 < 1,
10
Abstract and Applied Analysis
Table 6: Comparison of absolute errors for Problem (16).
𝑥
0.25
0.50
0.75
1.0
Present method
|𝑢 − 𝑢1 |
Akram and Rehman
[9], |𝑢 − 𝑢7 |
Golbabai and Javidi
[8]
(𝑁 = 7)
He [33]
Torvattanabun and
Koonprasert [34]
2.55351𝐸 − 15
9.65894𝐸 − 14
5.63438𝐸 − 13
9.23328𝐸 − 13
3.0291𝐸 − 10
7.7317𝐸 − 09
3.1222𝐸 − 08
4.3979𝐸 − 08
2.1630𝐸 − 09
1.1571𝐸 − 07
1.0479𝐸 − 06
4.2188𝐸 − 06
4.578𝐸 − 09
9.840𝐸 − 09
1.096𝐸 − 05
1.861𝐸 − 04
3.8922𝐸 − 10
1.1571𝐸 − 07
1.0479𝐸 − 06
4.2188𝐸 − 06
Table 7: Comparison of maximum absolute errors for Problem (17).
Present
method
Geng and Li
[10]
Siddiqi et al. [11]
Siddiqi and Akram
[12]
6.42 × 10−13
9.08 × 10−12
1.97 × 10−6
3.28 × 10−6
×10−13
6
4
2
−0.5
2.07 × 10−3
1.86 × 10−8
1.75 × 10−12
Following the procedure of the previous example, this
problem is solved using the convergence control parameter
as ℎ = −1.
It is observed that the errors in absolute values of the
present method are better than those of Farajeyan and Maleki
[19], Geng and Li [10], Lamnii et al. [14], Siddiqi and Akram
[12], Siddiqi et al. [11], and Siddiqi and Twizell [13] as shown
in Table 7. Figures 13 and 14 show the comparison of exact
solution with approximate solution and absolute errors for
Example 7, respectively.
8
−1
Siddiqi and Twizell
Farajeyan and Maleki
Lamnii et al. [14]
[13]
[19]
𝑥 ∈ [𝑥5 , 𝑥𝑘−5 ]
Example 8. Consider the following tenth-order nonlinear
boundary value problem:
1
0.5
Figure 14: Absolute errors for problem (17).
𝑢(10) (𝑥) =
Table 8: Comparison of numerical results for Example 8.
𝑥
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Exact solution
−0.0473684
−0.0888889
−0.123529
−0.15
−0.166667
−0.171429
−0.161538
−0.133333
−0.08181824
Absolute error
present method
Absolute error
Kasi Viswanadham
and Raju [35]
3.95413𝐸 − 11
7.33317𝐸 − 10
7.33317𝐸 − 09
6.06524𝐸 − 09
7.74775𝐸 − 09
6.56402𝐸 − 09
3.48667𝐸 − 09
9.23198𝐸 − 10
5.33521𝐸 − 11
1.322478𝐸 − 06
4.231930𝐸 − 06
1.676381𝐸 − 05
4.245341𝐸 − 05
6.663799𝐸 − 05
6.940961𝐸 − 05
4.750490𝐸 − 05
1.643598𝐸 − 05
2.607703𝐸 − 07
𝑢(3) (−1) = 0,
𝑢(4) (−1) =
−4
,
𝑒
𝑢(3) (1) = −12𝑒,
𝑢(3) (1) = −20𝑒.
(49)
The exact solution of problem (17) is 𝑢(𝑥) = (1 − 𝑥2 )𝑒𝑥 [10–
14, 19].
14175
(𝑥 + 𝑢 (𝑥) + 1)11 ,
4
𝑢 (0) = 0,
0 < 𝑥 < 1,
𝑢 (1) = 0,
−1
,
2
𝑢(1) (1) = 1,
1
𝑢(2) (0) = ,
2
𝑢(2) (1) = 4,
3
𝑢(3) (0) = ,
4
𝑢(3) (1) = 12,
3
𝑢(4) (0) = ,
2
𝑢(3) (1) = 48.
𝑢(1) (0) =
(50)
The exact solution of problem (18) is 𝑢(𝑥) = (2/(2 − 𝑥)) − 𝑥 − 1
[35].
Following the procedure of the previous example, this
problem is solved using the convergence control parameter
as ℎ = −1.
It is observed that the errors in absolute values are better
than those of Viswanadham and Raju [35] as shown in
Table 8. In Figure 15, the comparison of exact solution with
approximate solution is shown, and in Figure 16, absolute
errors for Example 8 are shown, respectively.
Abstract and Applied Analysis
0.2
0.4
11
0.6
0.8
1
−0.025
−0.05
−0.075
−0.1
−0.125
−0.15
−0.175
Figure 15: Comparison of the approximate solution with the exact
solution for problem (18). Dotted line: approximate solution; solid
line: the exact solution.
×10−9
6
4
2
0.2
0.4
0.6
0.8
1
Figure 16: Absolute errors for problem (18).
4. Conclusion
In this paper, the homotopy analysis method (HAM) has
been applied to obtain the numerical solutions of seventheighth-, and tenth-order boundary value problems. All computational work was carried out using Mathematica software.
The numerical results show that only a few number of
approximations can be used for numerical purpose with a
high degree of accuracy. It is observed that the absolute errors
are better than the methods in [3, 5–14, 19, 32–35]. It is
also observed that our proposed method is well suited for
the solution of higher order boundary value problems and
reduces the computational work. HAM converges to exact
solutions more rapidly as compared to the other method.
Therefore, the present method is an accurate and reliable
analytical technique for boundary value problems.
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